Theory Of Production - Returns To One Variable Factor

If the fixed volume of capital in the short run parities 1000 units, what is the
short run production function?

Depict the marginal product of labour MPL is less than average product of labour
AP in the short run production function in (2).

Solution

To obtain marginal product of labour, we distinguish the provided production function
with respect to labour. Therefore

MPL = dQ = 0.50L^-0.50
K^0.50
dL

= 0.50(K/L)
^ 0.50

Note that in the short run production function, one factor is variable with the
quantity of other fixed factor. Therefore,

Q = L^0.50
K ^0.50

Q = L^0.50
(900)^0.50

Since the square root of 900 is 30,

Q = L^0.50
* 30

= 30L^0.50

Therefore, short run production function is Q = 30L^0.50

MPL = dQ
dL

= 30
* 0.50L^-0.50

= 15
L^0.50

APL = Q
/ L

= 30L^0.50
L

= 30L^0.50-1

= 30
L^0.50

Relating the values of MPL and APL we determine that MPN < APL

Illustration 33

Let us assume the following production function of an industry,

O
= 3L^2 – 0.2L^3

Where O is output and L is the volume of Labour used.

Ascertain the MPL

Value of L that optimises output O

Value of L at which its Average Product is optimum

Solution

It is noted that the above function is short run production function as there is no
fixed term in it; all terms in it contain the variable factor, Labour (L).

(1) MPL = dO
dL

= 2
* 3L – 3 * 0.2L ^ 2

= 6L – 0.6L^2

To determine the average product AP of labour, we must divide the aggregate output by
L, therefore,

APL = O = 3L^2 – 0.2L^3
L L

= 3L – 0.2L^2

We can determine the value of the variable factor L that optimises output O and
also the value of labour L at which its average product is optimum.

The value of variable input L optimises output O that ca be procured by setting marginal
product function of the variable input equal to null. We have obtained above that MPL
= 6L – 0.6L^2. Setting it equal, to zero we have

6L – 0.6L^2 = 0

0.6L^2
= 6L

0.6L^2 = 6
L

0.6L = 6

L = 6
/ 0.6 = 10

At 10 units of labour, value of O will be optimum.

Value of L at which its AP is optimum:

Value of average product function will be optimised where its first derivative equals
zero.

AP of labour procured above = 3L – 0.2L^2

= dAP = 3 – 0.4L = 0
dL

0.4L = 3

L = 3
/ 0.4 = 7.5

Therefore, when 7.5 units of labour are used its average product will be optimum.

Illustration 34

Presume a firm producing cotton cloth has the following production function.

O = 4K^ ½ *
2L^ ½

Ascertain the marginal product of labour and capital.

Solution

This is Cobb-Douglas production function specific values of exponents.

MPL = dO
dL

= 4
* (½) K^ ½ * 2L ^ ½ – 1

= 2 √ K
* 2 -√L

= K
L

MPK = dO = 4
* (1/2) K^ ½ - 1 * 2 L^ ½
dK

= 2K^- ½ *
2 √ L

= 2L
K

Illustration 35

Let us assume the following production function

O
= 1.50 A^0.75 B^0.25

Determine the elasticity of productivity O with respect to A – Labour and B – Capital.
Provide an economic interpretation of this productivity elasticity.

Solution

Provided the production function is

O
= 1.50 A^0.75 B^0.25

Productivity elasticity of labour EA = MPA
APA

MPA = dO = 0.75
* 1.50L^-0.25 * B^0.25
dA

APA = O = 1.125A^0.75
* B^0.25
A

= 1.50A
^0.75 * B^0.25
A

= 1.50A
^-0.25 * B^0.25

EA = MPA = 1.125A^-0.25
* B^0.25
APA 1.50A
^-0.25 * B^0.25

= 0.75

Productivity elasticity of capital EB = MPB
APB

MPB = dO = 0.25
* 1.50 A^0.75 * B^-0.75
dB

= 0.375
L^0.75 * K^-0.75

APB = O = 1.50A^0.75
* B^0.25
B B

= 1.50A^0.75
* B^-0.75

EB = MPB = 0.375
A^0.75 * B^-0.75
APB 1.50A^0.75
* B^-0.75

= 0.25

From the value of elasticity of labour equal to 0.75 it follows
that 1 percent enhancement in employment of labour causes 0.75 enhancements in productivity,
which is less than ration. Likewise, productive elasticity of capital being equal to
0.25 entails that one percent enhancement in capital causes 0.25 percent enhancement
in productivity of the product.

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